Question

(3Sin alfa+ 3Cos alfa) в квадрате - 9Sin2 alfa

Answer 1

$<var>(3sina+3cosa)^{2}-9sin2a=9sin</var><var>a</var>^{2}+18sin<var>a</var>cos<var>a</var>+9cos<var>a</var>^{2}<var>-9sin2a</var>=9(<var>sin</var><var>a</var>^{2}<var></var>+cos<var>a</var>^{2}<var></var>)+9sin2a-9sin2a=9$